Prove that $\frac{x}{x+1}<\ln(1+x)-1,x\neq 0$ I found this problem in the mean value theorem section in a real analysis book.
I did not know, how to use the mean value theorem but I tried to find $f'$ 
and I found
$$\frac{1}{(x+1)^{2}}<\frac{1}{1+x}<1.$$
Clearly, $\frac{1}{(x+1)^{2}}$decrease faster than$\frac{1}{1+x}$, when $x$ grow larger.
What I will do next if I want to start from this, or is there another way to prove it? 
 A: Let $f(x) = \log(1+x)$ for $x > -1$.  Then $f'(x) = \frac{1}{1+x}$ and
the mean-value theorem states that for $x \ne 0$
$$
 f(x) = f(0) + (x-0) f'(c) = \frac{x}{1+c}
$$
for some $c$ strictly between $0$ and $x$.
If $x > 0$ then 
$$
 0 < c < x \Longrightarrow \frac{1}{1+x} < \frac{1}{1+c} < 1\, ,
$$
and if $x < 0$ then 
$$
 x < c < 0 \Longrightarrow \frac{1}{1+x} > \frac{1}{1+c} > 1\, .
$$
In both cases, multiplying the inequalities by $x$ gives
$$
\frac{x}{1+x} <  f(x) = \frac{x}{1+c} < x \, .
$$
A: Let $f(x)=x-\ln(1+x)$, where $x>-1$.
$f'(x)=1-\frac{1}{1+x}=\frac{x}{1+x}$, which says that $f(x)\geq f(0)=0$.
In another hand, let $g(x)=\ln(1+x)-\frac{x}{x+1}$, where $x>-1$.
Thus, $g'(x)=\frac{1}{1+x}-\frac{1}{(1+x)^2}=\frac{x}{(1+x)^2}$, which gives again 
that $x_{min}=0$ and we obtain $g(x)\geq g(0)=0$.
Since, $x\neq0$ we get $\frac{x}{x+1}<\ln(1+x)<x$.
Done!
A: Let's replace $x$ by $x-1$ to get $$\frac{x-1}{x}<\log x<x-1$$ for all $x>0,x\neq 1$. For $x=1$ the symbol $<$ needs to be replaced by $=$. Let's first assume the inequality $$\log x<x-1\tag{1}$$ which is more fundamental and we show that the other inequality is its consequence. Thus if we replace $x$ by $1/x$ in $(1)$ we get $$-\log x<\frac{1-x}{x}$$ or $$\frac{x-1}{x}<\log x$$ Thus it remains to prove the fundamental inequality $(1)$. This can be done by using any given definition of $\log x$. For example the easiest approach is to use the fact that $\log 1=0$ and $(\log x) '=1/x$. Then we can consider the function $f(x) =x-1-\log x$ and note that $f'(x) =1-(1/x)$ so that $f'(x) <0$ in $(0,1)$ and $f'(x) >0$ in $(1,\infty)$. It follows that $f(x) >f(1)=0$ for all $x>0,x\neq 1$ and we are done. 
